Let $f(x)=\dfrac{x^2}{e^x}$. $f'(x)=$
$f(x)$ is the quotient of two, more basic, expressions: $x^2$ and $e^x$. Therefore, the derivative of $f$ can be found using the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! = f ′ ( x ) = d d x ( x 2 e x ) = d d x ( x 2 ) e x − x 2 d d x ( e x ) ( e x ) 2 = 2 x ⋅ e x − x 2 ⋅ e x ( e x ) 2 = x e x ( 2 − x ) ( e x ) 2 = x ( 2 − x ) e x The quotient rule Differentiate x 2 and e x Simplify \begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}\left(\dfrac{x^2}{e^x}\right) \\\\ &=\dfrac{\dfrac{d}{dx}(x^2)e^x-x^2\dfrac{d}{dx}(e^x)}{(e^x)^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{2x\cdot e^x-x^2\cdot e^x}{(e^x)^2}&&\gray{\text{Differentiate }x^2\text{ and }e^x} \\\\ &=\dfrac{x\cancel{e^x}(2-x)}{(e^x)^\cancel2}&&\gray{\text{Simplify}} \\\\ &=\dfrac{x(2-x)}{e^x} \end{aligned} In conclusion, $f'(x)=\dfrac{x(2-x)}{e^x}$ or any other equivalent form.